All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.
Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.
Example:
Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
Output: ["AAAAACCCCC", "CCCCCAAAAA"]
所有的DNA都是由一系列核苷酸组成的，简写为 A, C, G, and T，比如："ACGAATTCCG"。当研究DNA时，识别DNA里的子序列是很有帮助的。写一个函数找出10个字母长度的出现过多次的子序列。
解法1：hash table + hash set
解法2: hash set
解法3：hash table + bit manipulte
Java:
public List findRepeatedDnaSequences(String s) {
Set result = new HashSet();
if(s ==null || s.length() <2)
return new ArrayList();
Set temp = new HashSet();
for(int i=0; i
Java:
public List findRepeatedDnaSequences(String s) {
Set seen = new HashSet(), repeated = new HashSet();
for (int i = 0; i + 9
Java: hashmap + bits manipulation
public List findRepeatedDnaSequences(String s) {
Set words = new HashSet<>();
Set doubleWords = new HashSet<>();
List rv = new ArrayList<>();
char[] map = new char[26];
//map['A' - 'A'] = 0;
map['C' - 'A'] = 1;
map['G' - 'A'] = 2;
map['T' - 'A'] = 3;
for(int i = 0; i
Python:
class Solution(object):
def findRepeatedDnaSequences(self, s):
"""
:type s: str
:rtype: List[str]
"""
dict, rolling_hash, res = {}, 0, []
for i in xrange(len(s)):
rolling_hash = ((rolling_hash <<3) & 0x3fffffff) | (ord(s[i]) & 7)
if rolling_hash not in dict:
dict[rolling_hash] = True
elif dict[rolling_hash]:
res.append(s[i - 9: i + 1])
dict[rolling_hash] = False
return res
Python:
def findRepeatedDnaSequences2(self, s):
"""
:type s: str
:rtype: List[str]
"""
l, r = [], []
if len(s) <10: return []
for i in range(len(s) - 9):
l.extend([s[i:i + 10]])
return [k for k, v in collections.Counter(l).items() if v > 1]
C++:
class Solution {
public:
vector findRepeatedDnaSequences(string s) {
unordered_set seen;
unordered_set dup;
vector result;
vector m(26);
m['A' - 'A'] = 0;
m['C' - 'A'] = 1;
m['G' - 'A'] = 2;
m['T' - 'A'] = 3;
for (int i = 0; i + 10 <= s.size(); ++i) {
string substr = s.substr(i, 10);
int v = 0;
for (int j = i; j
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